leetcode 题解代码整理 36-40题

Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character

'.'

.



A partially filled sudoku which is valid.
判断数独当前状态是否合法

class Solution
{
public:
bool isValidSudoku(vector<vector<char>>& board)
{
int col[10][10],row[10][10],box[10][10];
int i,j,x,a;
memset(col,0,sizeof(col));
memset(row,0,sizeof(row));
memset(box,0,sizeof(box));
for (i=0;i<9;i++)
for (j=0;j<9;j++)
if (board[i][j]!='.')
{
x=board[i][j]-'0';
a=i/3*3+j/3;
if (row[i][x]==1 || col[j][x]==1 || box[a][x]==1) return false;
else row[i][x]=col[j][x]=box[a][x]=1;
}

return true;
}
};

Sudoku Solver

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character

'.'

.

You may assume that there will be only one unique solution.



A sudoku puzzle...



...and its solution numbers marked in red.
DFS解数独，保证唯一解

class Solution
{
int col[10][10],row[10][10],box[10][10];

public:
void solveSudoku(vector<vector<char> >& board)
{
memset(col,0,sizeof(col));
memset(row,0,sizeof(row));
memset(box,0,sizeof(box));
for (int i=0;i<9;i++)
for (int j=0;j<9;j++)
if (board[i][j]!='.')
{
col[j][board[i][j]-'0']=1;
row[i][board[i][j]-'0']=1;
box[i/3*3+j/3][board[i][j]-'0']=1;
}

dfs(board,0);
}

bool dfs(vector<vector<char> > & board,int key)
{
int i,j;
if (key==81) return true;
int x=key/9;
int y=key%9;
if (board[x][y]!='.')
return dfs(board,key+1);
else
{
int a=x/3*3+y/3;
for (int i=1;i<=9;i++)
if (col[y][i]==0 && row[x][i]==0 && box[a][i]==0)
{
col[y][i]=row[x][i]=box[a][i]=1;
board[x][y]=i+'0';
if (dfs(board,key+1)) return true;
col[y][i]=row[x][i]=box[a][i]=0;
board[x][y]='.';
}
}
return false;

}
};

Count and Say

The count-and-say sequence is the sequence of integers beginning as follows:

1, 11, 21, 1211, 111221, ...

1

is read off as

"one
1"

or

11

.

11

is read off as

"two
1s"

or

21

.

21

is read off as

"one
2

, then

one 1"

or

1211

.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.
模拟过程

class Solution
{
public:
string countAndSay(int n)
{
vector<int>mark[n+1];
int temp;
char ch;
mark[1].push_back(1);
string ans;
int i,sum,j;

for (i=2;i<=n;i++)
{
temp=mark[i-1][0];
sum=1;
for (j=1;j<mark[i-1].size();j++)
if (mark[i-1][j]!=temp)
{
mark[i].push_back(sum);
mark[i].push_back(temp);
temp=mark[i-1][j];
sum=1;
}
else    sum++;

mark[i].push_back(sum);
mark[i].push_back(temp);
}
ans="";
for (i=0;i<mark
.size();i++)
{
ch=mark
[i]+'0';
ans=ans+ch;
}
return ans;

}
};

Combination Sum

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak).
The solution set must not contain duplicate combinations.

For example, given candidate set

2,3,6,7

and target

7

,

A solution set is:

[7]

[2, 2, 3]

输出序列中和=target的种类，可重复使用

Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where
the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2,
… , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤
… ≤ ak).
The solution set must not contain duplicate combinations.

For example, given candidate set

10,1,2,7,6,1,5

and target

8

,

A solution set is:

[1, 7]

[1, 2, 5]

[2, 6]

[1, 1, 6]

同上题，加一个条件，出现的数只能使用一次

class Solution
{
vector<vector<int> >ans;
vector<int>num;
int key;
int len;
private:
void dfs(vector<int>mark,int n,int sum,int used)
{

if (sum==key)
{
ans.push_back(mark);
return ;
}
if (n==len) return ;
if (sum+num
>key) return ;
dfs(mark,n+1,sum,0);
if (num
==num[n-1] && used==0) return ;

mark.push_back(num
);
dfs(mark,n+1,sum+num
,1);
mark.pop_back();

}
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target)
{
len=candidates.size();
num=candidates;
sort(num.begin(),num.end());
key=target;

vector<int>mark;
dfs(mark,0,0,1);
return ans;
}
};